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Unformatted text preview: garcia (jjg2564) HW 06 gualdani (56410) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the xintercept of the tangent line to the graph of f ( x ) = sin x + 3 cos x at the point (0 , f (0)). 1. xintercept = 3 4 2. xintercept = 3 correct 3. xintercept = 3 4. xintercept = 1 3 5. xintercept = 1 4 6. xintercept = 1 3 Explanation: When f ( x ) = sin x + 3 cos x , then f ( x ) = cos x 3 sin x . Thus at x = 0, f (0) = 3 , f (0) = 1 . So by the PointSlope formula, an equation for the tangent line at (0 , f (0)) is y 3 = 1( x 0) , which after rearranging become y = x + 3 . Consequently, the tangent line to the graph of f at (0 , f (0)) has xintercept = 3 . keywords: tangent, trig function, sin, cos, trig derivative, intercept, pointslope formula, 002 10.0 points Find the derivative of f when f ( x ) = sin x 2 + 3 cos x . 1. f ( x ) = 2 cos x 3 2 + 3 cos x 2. f ( x ) = 2 cos x 3 (2 + 3 cos x ) 2 3. f ( x ) = 2 cos x + 3 (2 + 3 cos x ) 2 correct 4. f ( x ) = 2 cos x + 3 2 + 3 cos x 5. f ( x ) = 2 cos x + 3 sin 2 x (2 + 3 cos x ) 2 Explanation: By the Quotient Rule, f ( x ) = cos x (2 + 3 cos x ) + 3 sin x sin x (2 + 3 cos x ) 2 = 2 cos x + 3(cos 2 x + sin 2 x ) (2 + 3 cos x ) 2 . But cos 2 x + sin 2 x = 1, so f ( x ) = 2 cos x + 3 (2 + 3 cos x ) 2 . 003 10.0 points Determine the derivative of f ( x ) = tan x 5 + sec x . 1. f ( x ) = tan x (5 sec x + 1) 5 + sec x garcia (jjg2564) HW 06 gualdani (56410) 2 2. f ( x ) = sec x (5 sec x 1) (5 + sec x ) 2 3. f ( x ) = 5 sec x tan x 5 + sec x 4. f ( x ) = tan x (5 sec x 1) 5 + sec x 5. f ( x ) = sec x (5 sec x + 1) (5 + sec x ) 2 correct 6. f ( x ) = 5 sec 2 x (5 + sec x ) 2 Explanation: Since (tan x ) = sec 2 x, (sec x ) = sec x tan x , the Quotient Rule ensures that f ( x ) = sec 2 x (5 + sec x ) sec x tan 2 x (5 + sec x ) 2 = 5 sec 2 x + sec x (sec 2 x tan 2 x ) (5 + sec x ) 2 . But sec 2 x = 1 + tan 2 x . Consequently, f ( x ) = sec x (5 sec x + 1) (5 + sec x ) 2 . 004 10.0 points Find an equation for the tangent line to the graph of f at the point P ( 4 , f ( 4 )) when f ( x ) = 5 tan x . 1. y = 15 x + 4 parenleftBig 1 2 parenrightBig 2. y = 10 x + 5 parenleftBig 1 2 parenrightBig correct 3. y = 2 x + 10 parenleftBig 1 4 parenrightBig 4. y = 18 x + 13 parenleftBig 1 2 parenrightBig 5. y = 9 x + 12 parenleftBig 1 4 parenrightBig Explanation: When x = 4 , then f ( x ) = 5, so P = ( 4 , 5)....
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This note was uploaded on 10/08/2010 for the course MATH 408K taught by Professor Gualdani during the Fall '09 term at University of Texas at Austin.
 Fall '09
 Gualdani

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